In triangle $\triangle ABC$, $AB = 37, AC = 33$, and $BC = 39$. A circle $\Gamma$ is drawn with center $A$ and radius $AC$. Let $D$ be the point on $\Gamma$ that is furthest from $B$. What is the distance $BD$?
BD = 33 + 37
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The greatest distance is just BA + AD = BD = 33 + 37 = 70